Answer:
He invested $2300 at 8%
Explanation:
* Lets explain how to solve the problem
- The student earned $5400
- He invested some of them at 8% and the rest at 6%
- He received from both a total $370 in interest at the end of the year
- The rule of the interest is I = Prt , where
# I is the interest
# P is the money invested
# r is the rat (in decimal)
* Lets change these information to equations
- Let the student invested $x at 8% and the rest at 6%
∵ The total investment in the two accounts is 5400
∵ He invested $x in the first account (8%)
∴ He invested $(5400 - x) in the second account (6%)
- Lets calculate the interest for each account I1 for first account of rate
8% and I2 for the second account of rat 6%
∵ I1 = Prt
∵ P = x and t = 1
∵ r = 8/100 = 0.08
∴ I1 = x(0.08)(1) = 0.08x
∵ I2 = Prt
∵ P = (5400 - x) and t = 1
∵ r = 6/100 = 0.06
∴ I2 = (5400 - x)(0.06)(1) = 5400(0.06) - x(0.06) = 324 - 0.06x
- The student received a total of $370 in interest at the end of the
year from both accounts
∴ I1 + I2 = 370
Substitute the values of I1 and I2 in the equation
∴ 0.08x + (324 - 0.06) = 370
- Collect the like terms
∴ (0.08 - 0.06) + 324 = 370
∴ 0.02x + 324 = 370
- Subtract 324 from both sides
∴ 0.02x = 46
- Divide both sides by 0.02
∴ x = 2300
∵ x is the money invested in the account of 8%
∴ He invested $2300 at 8%